<p> ST 361 HW1 Solutions</p><p>1. (a) Population: the used trucks sold in the car lot of the manager in last month. Sample: the 35 trucks sold last month. (b) variable: type of the truck Type of variable: categorical variable (c) Type of 1 2 3 4 5 6 Truck Proportion 9/35 6/35 6/35 4/35 6/35 4/35</p><p>(d) Four-wheel drive trucks: four-wheel drive Ford+ four-wheel drive Chevy + four-wheel Dodge = 6/35+4/35+4/35=2/5 (e) Pie-Chart</p><p>1(Ford) 2(4-Wheel Drive Ford) 3(Chevy) 4(4-Wheel Drive Chevy) 5(Dodge) 6(4-Wheel Drive Dodge)</p><p>Bar-Chart</p><p>3/10 1/4 1/5 3/20 1/10 Proportion 1/20 0 1(Ford) 2(4-Wheel 3(Chevy) 4(4-Wheel 5(Dodge) 6(4-Wheel Drive Ford) Drive Drive Chevy) Dodge) (f) The truck sold in Japan is a different population than what we have here, so we don’t have information of it.</p><p>1.15 (a) A histogram with classes of width 100 appears below. The histogram is positively skewed, and a representative data value is around 200. </p><p>Relative frequency</p><p>.30</p><p>.20</p><p>.10</p><p>.00 0 100 200 300 400 500 600 700 800 900 Number of cycles</p><p>2.2 (a)The sample mean is x = (100.4/8) = 12.55.</p><p>The sample size (n = 8) is even. Therefore, the sample median is the average of the (n/2) and (n/2) + 1 values. By sorting the 8 values in order, from smallest to largest: 8.0 8.9 11.0 12.0 13.0 14.5 15.0 18.0, the forth and fifth values are 12 and 13. The sample median is (12.0 + 13.0)/2 = 12.5.</p><p>All measures of center are similar, indicating little skewness to the data set.</p><p>(b) The smallest value (8.0) could be increased to any number below 12.0 (a change of less than 4.0) without affecting the value of the sample median.</p><p>(c) The values obtained in part (a) can be used directly. For example, the sample mean  1 ksi  of 12.55 psi could be re-expressed as (12.55 psi) x    5.70 ksi .  2.2 psi </p><p>2.3 (a) The display is reasonably symmetric, so the mean and median will be close. (b) The sample mean is x = 9638/26 = 370.7. The sample median is ~x = (369+370)/2 = 369.50.</p><p>(c) Expressed in minutes, the mean is (370.7 sec)/(60 sec) = 6.18 min; the median is 6.16 min.</p><p>2     2.19 Using the computational formula, s2 = 1  x 2  1  x   = [3,587,566- n1  i n  i    i  i   (9638)2/26]/(26-1) = 593.3415, so s = 24.36. In general, the size of a typical deviation from the sample mean (370.7) is about 24.4. Some observations may deviate from 370.7 by a little more than this, some by less.</p>